Example 1

For which primes $p$ is $p \cdot d(p) ≥ p^2$ ?

We must first note that $d(p) = 2$ for all primes $p$. Hence we need to find primes p such that $2p ≥ p^2$. We note that this is only true when $p = 2$ since $4 ≥ 4$. For all other primes, $2p < p^2$. Hence $p = 2$ is the only solution to this problem.

Example 2

For which primes $p$ and $q$ does $pq \cdot d(pq) ≥ p^2q^2$?

Since p and q are both primes, it follows that $(p, q) = 1$, and hence $d(pq) = d(p)d(q) = (2)(2) = 4$. Hence we want to find primes $p$ and $q$ that are true for the inequality $4pq ≥ p^2 q^2$.

Now fix $p = 3$. Hence we now want to solve the inequality that $12q ≥ 9q^2$, or rather $4q ≥ 3q^2$. But if $q$ is prime, then this is never true. In fact, $4pq ≥ p^2 q^2$ is ONLY true if $p = q = 2$. Hence we have only one solution.

Example 3

For which primes $p$ and $q$ does $d(p^q) > d(q^p)$.

We note that since $p$ and $q$ are both primes that $d(p^q) = q + 1$ and $d(q^p) = p + 1$. Hence we want to find values of $p$ and $q$ that satisfy $q + 1 > p + 1$, which reduces down to $q > p$. Hence for all primes q > p, $d(p^q) > d(p^q)$.

Example 4

Show that $d(n)$ is odd if and only if $n$ is a square.

First, suppose that $d(n)$ is odd. Since the positive divisors of $n$ come in pairs, this must mean that there exists a positive divisor $d$ that is repeated. Thus $n = d^2$ for some $d \in \mathbb{N}$.

Conversely, if $n$ is a square, then $n = d^2$ for some $d \in \mathbb{N}$. Suppose the prime power decomposition of $d = p_1^{e_1}p_2^{e_2}...p_k^{e_k}$. It thus follows that $n = p_1^{2e_1}p_2^{2e_2}...p_k^{2e_k}$. Hence: